Mechanical Waves ANSWERS TO QUESTIONS CHAPTER OUTLINE

Transcription

1 CHAPER OUNE. Proption o Disturnce. he We Model. he relin We.4 he Speed o rnserse Wes on Strins.5 Relection nd rnsmission o Wes.6 Rte o Enery rnser y Sinusoidl Wes on Strins.7 Sound Wes.8 he Doppler Eect.9 Context Connection Seismic Wes Mechnicl Wes ANSWERS O QUESONS Q. o use slinky to crete lonitudinl we, pull ew coils ck nd relese. or trnserse we, jostle the end coil side to side. Q. rom, we must increse the tension y ctor o 4. Q. t depends on rom wht the we relects. relectin rom less dense strin, the relected prt o the we will e riht side up. Q.4 he section o rope moes up nd down in SHM. ts speed is lwys chnin. he we continues on with constnt speed in one direction, settin urther sections o the rope into up-nd-down motion. Q.5 As the source requency is douled, the speed o wes on the strin stys constnt nd the welenth is reduced y one hl. Q.6 As the source requency is douled, the speed o wes on the strin stys constnt. Q.7 Hiher tension mkes we speed hiher. Greter liner density mkes the we moe more slowly. Q.8 As the we psses rom the mssie strin to the less mssie strin, the we speed will increse ccordin to. he requency will remin unchned. Since λ, the welenth must increse. Q.9 Amplitude is incresed y ctor o. he we speed does not chne. Q. Sound wes re lonitudinl ecuse elements o the medium prcels o ir moe prllel nd ntiprllel to the direction o we motion. 6

2 6 Mechnicl Wes Q. We ssume tht perect cuum surrounds the clock. he sound wes require medium or them to trel to your er. he hmmer on the lrm will strike the ell, nd the irtion will spred s sound wes throuh the ody o the clock. one o your skull were in contct with the clock, you would her the ell. Howeer, in the sence o surroundin medium like ir or wter, no sound cn e rdited wy. A lrer-scle exmple o the sme eect: Colossl storms rin on the Sun re dethly still or us. Wht hppens to the sound enery within the clock? Here is the nswer: As the sound we trels throuh the steel nd plstic, trersin joints nd oin round corners, its enery is conerted into dditionl internl enery, risin the temperture o the mterils. Ater the sound hs died wy, the clock will low ery slihtly rihter in the inrred portion o the electromnetic spectrum. Q. he requency increses y ctor o ecuse the we speed, which is dependent only on the medium throuh which the we trels, remins constnt. Q. When listenin, you re pproximtely the sme distnce rom ll o the memers o the roup. dierent requencies treled t dierent speeds, then you miht her the hiher pitched requencies eore you herd the lower ones produced t the sme time. Althouh it miht e interestin to think tht ech listener herd his or her own personl perormnce dependin on where they were seted, time l like this could mke Beethoen sont sound s i it were written y Chrles es. Q.4 He sw the irst we he encountered, liht trelin t 8. m s. At the sme moment, inrred s well s isile liht en wrmin his skin, ut some time ws required to rise the temperture o the outer skin lyers eore he noticed it. he meteor produced compressionl wes in the ir nd in the round. he we in the round, which cn e clled either sound or seismic we, treled much ster thn the we in ir, since the round is much stier inst compression. Our witness receied it next nd noticed it s little erthquke. He ws no dout unle to distinuish the P nd S wes. he irst ir-compression we he receied ws shock we with n mplitude on the order o meters. t trnsported him o his doorstep. hen he could her some dditionl direct sound, relected sound, nd perhps the sound o the llin trees. Q.5 or the sound rom source not to shit in requency, the rdil elocity o the source reltie to the oserer must e zero; tht is, the source must not e moin towrd or wy rom the oserer. he source cn e moin in plne perpendiculr to the line etween it nd the oserer. Other possiilities: he source nd oserer miht oth he zero elocity. hey miht he equl elocities reltie to the medium. he source miht e moin round the oserer on sphere o constnt rdius. Een i the source speeds up on the sphere, slows down, or stops, the requency herd will e equl to the requency emitted y the source. Q.6 Wind cn chne Doppler shit ut cnnot cuse one. Both o nd s in our equtions must e interpreted s speeds o oserer nd source reltie to the ir. source nd oserer re moin reltie to ech other, the oserer will her one shited requency in still ir nd dierent shited requency i wind is lowin. the distnce etween source nd oserer is constnt, there will neer e Doppler shit.

3 Chpter 6 Q.7 et t t t represent the dierence in rril times o the two wes t sttion t distnce s p d sts ptp rom the hypocenter. hen d t s p. Knowin the distnce rom the irst sttion plces the hypocenter on sphere round it. A mesurement rom second sttion limits it to nother sphere, which intersects with the irst in circle. Dt rom third non-colliner sttion will enerlly limit the possiilities to point. SOUONS O PROBEMS Section. Proption o Disturnce P. Replce x y x t x45. t to et y 6 x 45. t + P. G. P. Section. he We Model Section. he relin We 4. irtions 4 45 cm P. Hz s. s 4. 5 cm s λ. 9 cm. 9 m Hz 4 cm s

4 64 Mechnicl Wes P.4 Usin dt rom the osertions, we he λ. m nd 8.. s. hereore, 8. H G λ. m K J. 8 ms.. s π P.5 () et u πt π x+ 4 du dt dx dt dx π π t point o constnt phse dt. m s he elocity is in the positie x -direction. K J () y.,. 5 m sin. π + π 548. m 548. cm 4 π (c) k π : λ 667. m ω π π : 5. Hz λ (d) y y t x t 5 + π. π cos π π 4 in S units A P.6 y. m sin. x6. t k. rd m π λ k ω 6. rd s ω π ω π 6. λ π k. 7. m s P.7 () ω π π 5 s 4 rd s e j. ms () λ 4. m 5 s K J y, π. 5.. cm mx ms 98. m 576. Hz π π k 57. rd m λ 4 m (c) n y Asinkx ωt+ φ we tke A cm. At x nd t we he y cmsinφ. o mke this it y, we tke φ. hen y. cm sin. 57 rd m x. 4 rd s t d i (d) (e) he trnserse elocity is y Aωcoskxωt. ts mximum mnitude is t. Aω cm.4 rd s 77 m s y d i. m e. 4 s j 8 m s. y Aωcos kx ωt Aω sin kxωt t t he mximum lue is Aω

5 t coordinte x is 57. rd cm x5. rd st. Since 6 *P.8 At time t, the phse o y 5. cm cos. 57x5. t φ φ B π φ ± rd, or (since x A ), A. π rd, the requirement or point B is tht. 57 rd cm xb 5. rd s t 5. rd s t± π rd. Chpter 65 ± π rd his reduces to x B ± 667. cm 57. rd cm. π π P.9 () A ymx 8. cm 8. m k 785. m λ 8. m.. rd s sin ω. 8 sin π m ω π π 6π hereore, y A kx+ t Or (where y, t t t ) y x+ t () n enerl, y. 8 sin 7. 85x+ 6πt+ φ Assumin yx, t x. m then we require tht. 8 sin φ or φ 785. hereore, y. 8 sin 7. 85x+ 6π t. 785 m π P. y. m sin x+ 4π t 8 () K J dy y dt : y π. 4π cos 8 x+ 4π t y. s,.6 m. 5 m s d y : y x t dt π. m 4π sin 8 + 4π y. s,.6 m K J K J π () k 8 π λ : λ 6. m π ω 4π :. 5 s λ 6. m.5 s. ms

6 66 Mechnicl Wes P. () et us write the we unction s yx, t Asin kx+ ω t+ φ y, Asin φ. m dy dt, Aωcos φ. ms Also, π π ω 8. π s 5. s A i xi + K J +. ms. m 8 K J ω. π s A. 5 m () A sinφ. 5. tnφ A cosφ 8. π Your clcultor s nswer tn rd hs netie sine nd positie cosine, just the reerse o wht is required. You must look eyond your clcultor to ind φ π 9. rd 95. rd (c) y, mx Aω 5. m 8. π s 54. m s (d) λ x. ms. 5 s. 75 m π π k λ 75. m 8.8 m ω 8. π s P. he liner we eqution is yx, t. 5 m sin 8. 8 xrd m + 8. π trd s rd y e then hereore, y y x t xt y t e nd y xt y e t y t xt x e x t nd y e x xt y, demonstrtin tht x e t is solution x

7 Section.4 he Speed o rnserse Wes on Strins Chpter 67 P. he down nd ck distnce is 4. m+ 4. m 8. m. he speed is then dtotl 48. m 4. ms t 8. s. k Now, 5. k m 4. m e j So 5. k m 4. m s 8. N M M P.4 M is the tension; m m t is the we speed. hen, nd M m t e e j j m Mt. 6 m 4. k. k.6 s 64. ms *P.5 Since is constnt, nd P.6 rom the ree-ody dirm m sinθ. ms. ms H G K J H G K J m sinθ 8 he nle θ is ound rom cosθ () θ 4. 4 or. 4 ms m k 4 6. N 5. N. m 98. ms sin k m sin 4. 4 H G K J () m nd m 89. k e j r m r r G. P.6 m

8 68 Mechnicl Wes P.7 he totl time is the sum o the two times. n ech wire t et A represent the cross-sectionl re o one wire. he mss o one wire cn e written oth s m ρv ρ A nd lso s m. πρd hen we he ρa 4 hus, t or copper, t or steel, t πρ d 4 H G K J 89.. π e j 4 5 NM π e j 4 5 NM he totl time is s O QP O QP. 7 s 9. s Section.5 Relection nd rnsmission o Wes P.8 () the end is ixed, there is inersion o the pulse upon relection. hus, when they meet, they cncel nd the mplitude is zero. () the end is ree, there is no inersion on relection. When they meet, the mplitude is A. 5 m. m. Section.6 Rte o Enery rnser y Sinusoidl Wes on Strins. P.9 6. Hz ω π π rd s λ 5. P H G K J 8. ω A π... 7 kw 6.

9 Chpter 69 P.. m. k m λ 5. m 5. Hz : ω π 4 s A. 5 m : A 7. 5 m K J e j () y Asin π xω t λ y 7. 5 sin 4. 9x4t G. P. () P H G ω A e. j e. j K J W P 65 W P. A 5. m 4. k m P W N hereore, P ω A : ω 5. m s e je j P A ω 46 rd s ω 55. Hz π *P. Oriinlly, P ω A P ω A P ω A he douled strin will he douled mss-per-lenth. Presumin tht we hold tension constnt, it cn crry power lrer y times. P ω A Section.7 Sound Wes P. Since >> : d 4 ms 6. s km liht sound

10 7 Mechnicl Wes *P.4 he sound pulse must trel 5 m eore relection nd 5 m ter relection. We he d t d m t 5 m s. 96 s 4 ms P.5 () λ 48s. m 4 ms () λ 46. m, λ λ λ. 8 mm 97 s 4 ms P.6 λ 6. s 567. mm *P.7 he sound speed is ms + 6 ms C 6 C 47 ms. () et t represent the time or the echo to return. hen d t 47 ms 4 s 4. 6 m. () et t represent the durtion o the pulse: λ λ t λ 6 s 455. s. (c). 47 ms λ 6 s 58mm 5 ms *P.8 (). 4MHz, λ 6 4. s. 65 mm 5 ms () MHz, λ 5. mm 6 s 5 ms MHz, λ 75. m 7 s P.9 P ρωs ρ π H G K J s λ 6 s π.. mx e j 58. m mx mx mx πρ λ P 84. mx

11 Chpter 7 P. () A. m π λ. 4 m 4. cm 5. 7 ω 858 k m s e j m () s. cos e j (c) mx Aω. m 858s 7. mm s π π P. k 6. 8 m λ. m. e j sin 6. 8 m. 6 4 s. π π 4 ms ω 6. 4 s λ. m hereore, P x t 8 e j e j e j P. Pmx ρω smx. k m π s 4 m s. m P mx. P Section.8 he Doppler Eect P. () + 5 o s () 5 4 ( 4. ) (c) 5 4. khz.8 khz..6 khz while police cr oertkes khz ter police cr psses

12 7 Mechnicl Wes H G. rd s A e j P.4 () ω π 5 min π 6. smin mx ω. rd s. 8 m. 7 m s () (c) he hert wll is moin oserer. O + K J. + Now the hert wll is moin source. 5 7 Hz 8. 9 Hz Hz Hz s 5 7 P.5 Approchin mulnce: Deprtin mulnce: Since 56 Hz nd S S d i S K J + 48 Hz S S S P.6 he mximum speed o the speker is descried y ms ms mmx ka k mx 5 m A. Nm. m. m s 5. k he requencies herd y the sttionry oserer rne rom min + mx to mx mx K J where is the speed o sound. 44 Hz min 44 Hz mx 4 m s 4 ms +. ms 4 m s 4 ms. ms 49 Hz 44 Hz

13 P.7 d s t ll + d. t i K J t s 8. t 8. m: t return 58. s 4 he ork continues to ll while the sound returns. m P.8 () ms +. 6 C 5 m s s C ttotl ll t + treturn. 9 s s. 985 s dtotl ttotl ll 9. m () Approchin the ell, the thlete hers requency o Ater pssin the ell, she hers lower requency o he rtio is + + O K J + O O O 5 6 Chpter 7 which ies 6 6o 5+ 5o or O 5 ms 9. 5 ms P.9 () Sound moes upwind with speed 4 5 m s. Crests pss sttionry upwind point t requency 9 Hz. hen 8 ms λ 9 s 64. m () By similr loic, λ ms 9 s 98. m (c) (d) he source is moin throuh the ir t 5 m/s towrd the oserer. he oserer is sttionry reltie to the ir. + o Hz K J 94 Hz s 4 5 he source is moin throuh the ir t 5 m/s wy rom the downwind ireihter. Her speed reltie to the ir is m/s towrd the source. H G K J + o Hz 9 Hz 98 Hz s

14 74 Mechnicl Wes Section.9 Context Connection Seismic Wes P.4 () he lonitudinl we trels shorter distnce nd is moin ster, so it will rrie t point B irst. () he we tht trels throuh the Erth must trel e j 6 6 distnce o R sin m sin m t speed o 7 8 m/s hereore, it tkes m 87 s 7 8 m s he we tht trels lon the Erth s surce must trel distnce o s R RH G π θ rd K J m t speed o hereore, it tkes 4 5 m/s s he time dierence is 665 s. min P.4 he distnce the wes he treled is d 78. km s t 45. km s t+ 7. s where t is the trel time or the ster we. 45. km s 7. s. 6 s km s 78. km s 6. s 84km. hen, km s t 45. km s 7. s or t nd the distnce is d Additionl Prolems P.4 Assume typicl distnce etween djcent people ~ m. x hen the we speed is t ~ m ~ ms. s Model the stdium s circle with rdius o order m. hen, the time or one circuit round the stdium is r π πej ~ 6 s ~ min. ms

15 Chpter 75 P.4 Assumin the incline to e rictionless nd tkin the positie x-direction to e up the incline: x Msinθ or the tension in the strin is M sinθ he speed o trnserse wes in the strin is then M sinθ m M sinθ m m he time interl or pulse to trel the strin s lenth is t M sinθ m M sinθ P.44 Mx kx () kx M M () + x + k (c) M m m M + k *P.45 et M mss o lock, m mss o strin. or the lock, m implies m mω r. he r speed o we on the strin is then K J M r ω M rω m r m r m t ω M m. k θ ωt M.45 k 84. rd P.46 () Assume the sprin is oriinlly sttionry throuhout, extended to he lenth much reter thn its equilirium lenth. We strt moin one end orwrd with the speed t which we proptes on the sprin. n this wy we crete sinle pulse o compression tht moes down the lenth o the sprin. or n increment o sprin with lenth dx nd k mss dm, just s the pulse swllows it up, m ecomes kdx dm or dm dx () dm dx so k. Also, d dt t when. But t, so i expressions or, we he k k or.. Equtin the two k k Nm. m Usin the expression rom prt () m 4. k. 6 ms.

16 76 Mechnicl Wes P.47 where x, the weiht o lenth x, o rope. hereore, x dx dx But, so tht dt dt x nd dx x t x z mx P.48 At distnce x rom the ottom, the tension is M H G K J +, so the we speed is: z z NM () hen t dt x M dx x m + K J m. dt O QP t M + H G m K J NM dx M M t + m K J H G m K J t x + M m O t QP m m K J m m m M M K J + + K M 8 M e j 8 e j K m m M () When M, s in the preious prolem, t (c) As m we expnd m+ M M + to otin t H G t M + m M m M + M x ω P.49 () P x A A e H G ω ω ω k K J k A e () P (c) ω k A P x P e x 445km P km h m s 9.5 h e j ms d 98. ms 7 m m M x x x m+ M M m

17 P.5 () x is liner unction, so it is o the orm x mx+ o he so we require. hen m+ hen x m x + Chpter 77 () dx rom, the time required to moe rom x to x+ dx is dx. he time required to moe dt rom to is dx dx t t z z z z x + xdx dx K J x t + t e j e je + + j t t P.5 Sound tkes this time to rech the mn: e j e j. m. 75 m 4 ms 5. s so the wrnin should e shouted no lter thn. s+ 5. s. 5 s eore the pot strikes. Since the whole time o ll is ien y y t : m e. m s jt t 9. s the wrnin needs to come. 9 s. 5 s. 58 s into the ll, when the pot hs llen e j 9. 8 ms. 58 s. m to e oe the round y. m. m 7. 8 m P.5 Since cos θ + sin θ, sinθ ± cos θ (ech sin pplyin hl the time) P P sin kx ωt ± ρωs cos kxωt mx mx mx mx mx hereore P± ρω s s cos kx ωt ± ρω s s

18 78 Mechnicl Wes P.54 he trucks orm trin nloous to we trin o crests with speed 9. 7 m s nd unshited requency 667. min.. min () he cyclist s oserer mesures lower Doppler-shited requency: () + o K J e + o K J e 667. min. 667 min j j min 64. min he cyclist s speed hs decresed ery siniicntly, ut there is only modest increse in the requency o trucks pssin him. e j d P.55 t : d 65. ms 85. s 6. km t P.56 () u u + + u u e j u u u u + u 6. 4 () km h 6. m s Hz 4 6. *P.57 () dier dier so dier with 4 m s, 8 Hz nd 5 Hz u u 4 K J we ind dier 4 8 K J ms. 5 () the wes re relected, nd the skydier is moin into them, we he so dier 5Hz. NM dier O QP + dier

19 Chpter 79 *P.58 () G. P.58() () 4 ms λ s 4. m (c) S λ K J 4 4. ms s. m (d) + S ms λ s. 8 m (e) K J O 4 Hz. S 4 4. ms ms. khz *P.59 Use the Doppler ormul, nd rememer tht the t is moin source. the elocity o the insect is x, x x Solin, x. m s. hereore, the t is inin on its prey t.69 m s. P.6 () the source nd the oserer re moin wy rom ech other, we he: θ S θ 8, nd since cos8, we et Eqution. with netie lues or oth O nd S. () O m s then. Also, when the trin is 4. m rom the intersection, S cosθ S nd the cr is. m rom the intersection, cosθ S 4 5 so 4 ms 5 Hz 4 ms. 85. ms or 5 Hz. Note tht s the trin pproches, psses, nd deprts rom the intersection, θ S ries rom to 8 nd the requency herd y the oserer ries rom: 4 ms mx 5 Hz 59 Hz S cos 4 ms 5. ms 4 ms min cos 8 Hz Hz 4 ms + 5. ms S

20 8 Mechnicl Wes ANSWERS O EVEN PROBEMS P. see the solution P.4 8. m s P.6. cm,.98 m,.576 Hz,.7 m/s P.8 ±667. cm P. ().5 m/s, m s ; () 6. m,.5 s,. m/s P. see the solution P m s P.6 (). 4 ms m ; ().89 k k H G K J P.8 () zero; (). m e j ; P. () y 7. 5 sin 4. 9x4t () 65 W P. P P.4.96 s P mm P.8 ().65 mm; ().5 mm to 75. m P. (). m, 4. cm, 54.6 m/s; (). 4 m; (c).7 mm/s P.. P P.4 (). 7 m/s; () 8.9 Hz; (c) 57.8 Hz P.6 49 Hz nd 44 Hz P.8 () 5 m/s; () 9.5 m/s P.4 () lonitudinl ; () 665 s P.4 ~ min M P.44 () M ; () + ; k M (c) m M + k K J P.46 () P.48 see the solution P.5 m/s, 7m P m k ; ().6 m/s P.54 () 55. min; () 64. min P.56 () u u ; () 85.9 Hz P.58 () see the solution; ().4 m; (c). m; (d).8 m; (e). khz P.6 () see the solution; () 5 Hz